\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) 
           2,25     1,5  
=> \(V_{O_2}=1,5.22,4=33,6\left(L\right)\) 
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\) 
                1                             1,5 
=> \(m_{KClO3}=122,5\left(g\right)\)

\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)

\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\\ pthh:2H_2O\underrightarrow{\text{đ}p}2H_2+O_2\) 
          0,2         0,2      0,1 
=> \(\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(L\right)\\V_{O_2}=0,1.22,4=2,24\left(L\right)\end{matrix}\right.\)

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\\ pthh:2K+2H_2O->2KOH+H_2\) 
          0,2       0,2        0,2               0,1
=> \(V_{H_2}=0,1.22,4=2,24\left(L\right)\) 
     \(m_{H_2O}=0,2.18=3,6\left(g\right)\\ m_{KOH}=0,2.56=11,2\left(g\right)\) 

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) 
           0,5                   0,25                     0,25 
=> \(\left\{{}\begin{matrix}m_{KMnO_4}=0,5.158=79\left(mol\right)\\m_{K_2MnO_4}=0,25.197=49,25\left(g\right)\end{matrix}\right.\)

\(n_{CuO}=\dfrac{160}{80}=2\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

              2 <---- 1 <-------- 2

\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=1.22,4=22,4\left(l\right)\\m_{Cu}=2.64=128\left(g\right)\end{matrix}\right.\)

nCuO=160/80=2mol

2Cu+O2→(to)2CuO

 2            1           2          ( mol )

a)VH2=1.22,4=22,4l

b)mCu=2.64=128g

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

LTL: \(\dfrac{1,5}{2}< 1,5\rightarrow O_2\) dư

Theo pt: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.1,5=0,75\left(mol\right)\\n_{H_2O}=n_{H_2}=1,5\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(1,5-0,75\right).32=24\left(g\right)\\V_{O_2}\left(1,5-0,75\right).22,4=16,8\left(l\right)\\m_{H_2O}=1,5.18=27\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(MOL\right)\) 
pthh: \(2H_2+O_2\underrightarrow{t^O}2H_2O\) 
LTL : \(\dfrac{1,5}{2}< \dfrac{1,5}{1}\) 
=> O2 dư , H2 hết 
theo pthh: nH2O = nH2 = 1,5 (mol) 
 => \(m_{H_2O}=1,5.18=27\left(g\right)\)